Problem 30 - Calculus Early Transcendentals

Question

At what point does the curve have maximum curvature? What happens to the curvature as \(x \rightarrow \infty ?\) \(y=\ln x\)

Step-by-Step Solution

Answer

Maximum curvature occurs near \(x \approx 1\), and curvature approaches 0 as \(x \rightarrow \infty\).

Step 1: Understanding Curvature

Curvature of a function is a measure of how quickly it changes direction. For a curve defined by a function, the curvature is given by: \[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\]where \(y'\) and \(y''\) are the first and second derivatives of the function.

Step 2: Find the First Derivative

Determine the first derivative of the function \(y = \ln x\) using standard differentiation rules.\[y' = \frac{d}{dx}\ln x = \frac{1}{x}\]

Step 3: Find the Second Derivative

Differentiate \(y' = \frac{1}{x}\) to find the second derivative.\[y'' = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}\]

Step 4: Substitute into Curvature Formula

Substitute \(y'\) and \(y''\) into the curvature formula:\[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}} = \frac{|-\frac{1}{x^2}|}{(1 + \left(\frac{1}{x}\right)^2)^{3/2}}\]

Step 5: Simplify Curvature Expression

Simplify the expression for curvature:\[\kappa = \frac{\frac{1}{x^2}}{(1 + \frac{1}{x^2})^{3/2}} = \frac{1}{x^2(1 + \frac{1}{x^2})^{3/2}}\]

Step 6: Analyze Curvature for Maximum

To find the maximum curvature, analyze the behavior of \(\kappa\). As \(x\rightarrow 0^+\), \(y(x)\rightarrow -\infty\), which isn't feasible in the context of the function's domain (\(x > 0\)). As \(x \rightarrow \infty\), the expression inside the denominator approaches \(x^3\), and \(\kappa = \frac{1}{x^2} \cdot \frac{x^{3/2}}{x^3} = \frac{1}{x} \rightarrow 0\). The maximum curvature occurs near the smallest positive \(x\), which happens when changes in direction are greatest, often near the general range where it is defined prominently close to 1.

Step 7: Conclusion about Limiting Behavior

As \(x \rightarrow \infty\), the term \(\frac{1}{x}\) causes the curvature, \(\kappa\), to approach zero, meaning the curve flattens out with increasing \(x\).

Key Concepts

Curvature formula

Understanding the concept of curvature is essential when dealing with curves. Curvature provides a quantitative measure of how much a curve deviates from being a straight line. The formula used to calculate the curvature \[\kappa = \frac{|y''|}{(1 + (y')^2)^{3/2}}\] uses the first and second derivatives of a function. These derivatives show how a function's rate of change varies, with the second derivative describing the change in rate of change.
Thus, to find the curvature of any function at a given point, it's crucial to first find these derivatives. Applying this formula to various functions allows you to determine points of maximum bending or turning on a curve.

• Numerator: The absolute value of the second derivative, \(|y''|\), shows how sharply the curve bends.
• Denominator: The expression \((1 + (y')^2)^{3/2}\) normalizes the curvature, incorporating the first derivative \(y'\) to ensure the curvature is correctly scaled.

First derivative

The first derivative of a function, often denoted as \(y'\), is a fundamental concept in calculus. It represents the rate at which the function value changes as the input changes. For the function \(y = \ln x\), the first derivative is computed using differentiation rules.
When differentiating the logarithmic function, the first derivative is \[ y' = \frac{1}{x} \]which indicates how quickly the function changes at each point \(x\).

• The first derivative helps in finding the slope of the tangent line to the curve at any given point.
• It is crucial in identifying increasing or decreasing behavior of the function across its domain.

Second derivative

The second derivative of a function, typically denoted as \(y''\), provides more detailed information about the curve. Specifically, it tells us about the concavity of the curve and changes in the rate of change (acceleration).
For the function \(y = \ln x\), the second derivative is given by differentiating the first derivative: \[ y'' = -\frac{1}{x^2} \]This negative second derivative indicates a downward concavity across the domain (the curve is concave down).

• When \(y'' > 0\), the curve is concave up, and when \(y'' < 0\), it is concave down.
• In curve analysis, the second derivative is essential for understanding points of inflection and curvature.

Curve analysis

Curve analysis involves evaluating the behavior of a function by studying its derivatives. This process reveals crucial aspects of how a curve behaves, including identifying points that are peaks, valleys, or points of inflection.
Given the function \(y = \ln x\), we can conduct a thorough analysis using its first and second derivatives. With \(y' = \frac{1}{x}\ \)and \(y'' = -\frac{1}{x^2} \), we understand that the curve continuously decreases in steepness for all \(x > 0\).

• Using the derivatives helps in sketching/predicting the shape of the graph.
• Important for identifying both local and global maxima and minima of the function, although in this specific case, it reveals no local extrema.

Limit behavior

The limit behavior of a function or its derivatives describes what happens as the variable approaches a particular point or infinity. Understanding this behavior is essential, especially for functions that continue indefinitely in their domains.
As \(x \rightarrow \infty\) with the function \(y = \ln x\), the curvature \[\kappa = \frac{1}{x^2 (1 + \frac{1}{x^2})^{3/2}} \]trends towards zero. This result tells us the curve increasingly resembles a straight line, flattening out as it extends to larger \(x\) values.

• The limit behavior helps in understanding the end behavior of functions.
• It assists in simplifying complex expressions and indicating trends or tendencies over various domains.